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Solve the equation on the interval [0,2 pie) 2 cos 2x - square root of 3 = 0
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2cos2x-sqrt3 = 0
`cos2x = sqrt3/2`
`cos2x = cos(pi/6)`
`2x = 2npi+-pi/6`
`x = npi+-pi/12`
When n = 0 then `x = -pi/12` and `x = pi/12`
When n = 1 then `x = (11pi)/12` and `x = (13pi)/12`
When n = 2 then `x = (23pi)/12` and `x = (25pi)/12`
So the answers between 0 and `2pi ` are;`x = pi/12,(11pi)/12,(13pi)/12,(23pi)/12`
Posted by jeew-m on July 17, 2013 at 12:18 AM (Answer #1)
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